H2 maths tutor needs help
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Hi,
Questions on statistics are usually tricky because of the play of words. Students will do better if we coach them on skills to interpret the meaning of sentences correctly. My students give me the impression that schools focus only on content rather than metacognitive strategies. Thanks.
Cheers,
Wen Shih -
Actually I really wonder about how schools teach maths. It seems that they have the ability to turn simple concepts into amazingly difficult ones for the students. It doesn't help when students have self-defeating mentalities that it is hard as well.
Btw, sanity checks (or reasonableness checks) are common place in the field of engineering. Guess I was already part engineer mindset during my JC days with all these quick checks
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That's quite enlightening about the engineering mind-set.
I indeed had a bad lecturer in JC who made everything seem harder than it was. My current student also suffers because her teacher is too absorbed with complex formulae. When she shows me her lecture notes, I often cringe. They are overloaded with profound-looking formulae which intimidates students unnecessarily!
Is there a movement in Singapore devoted to improving the teaching of H2 maths?
Also, speaking of stats wording, can help me out with this, which I encountered this week? This is on my student's tutorial and supposedly from J79/P2/13 (Modified):
In a certain population, the heights of men are normally distributed with mean 172cm and standard deviation 10cm, and the heights of women are normally distributed with mean 165cm and standard deviation 8cm.
Calculate the probability that the height of a randomly chosen man exceeds that of a randomly chosen woman by 10cm.We are given M ~ N(172, 102) and W ~ N(165, 82).
Okay, so we can define a new variable X = M - W.
Then X ~ N(7, 164).But what is the question asking? It seems to be asking for P(X = 10), but that cannot be, since the answer would just be 0.
Maybe it is really asking for P(X > 10), in which case the wording should have been this:
Calculate the probability that the height of a randomly chosen man exceeds that of a randomly chosen woman by more than 10cm.
But when I calculate the answer, I get 0.4075, whereas the given answer is 0.309. So I don't really know what this question is asking. Any ideas?
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is your student from RVHS?
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Originally posted by Mad Hat:
Hi Mr. Wee, thanks for confirming that question! I have another similar question from the TYS which I will post soon. As for checking for typos, you're welcome. I will PM you a list before long - I'll check the latest DYNA edition first in case some of the typos have already been corrected.
In appreciation for all the help, may I share with everyone the following probability problems, which are simple yet enlightening:
This is a famous trick question in conditional probability. The natural answer is 1/2 but, surprisingly, the correct answer is 1/3! This is within the H2 syllabus.
But now consider this question:
This looks like the same question but, surprisingly, in this case, the answer of 1/2 is perfectly correct! Am I right?
Thankfully, the difference between the two questions is not within the H2 syllabus, but I'll be happy to discuss both problems if anyone is interested.
this is actually a re-incarnation of the famous Monty Hall problem.
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
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Originally posted by SBS261P:
this is actually a re-incarnation of the famous Monty Hall problem.
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
If it was me, I will keep with No. 1.
It seems as if the host is trying to play mind games with me and make me change my choice.
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Originally posted by Forbiddensinner:
If it was me, I will keep with No. 1.
It seems as if the host is trying to play mind games with me and make me change my choice.
no, by keeping to your choice you have only a 1/3 chance of winning the car. -
Dear Mad Hat,
We can always start a movement to help students understand H2 maths on this forum :)
Sometimes, simple things are made complex because of a fixated mindset that concepts must be taught in a certain way, rather than by the use of skilful means that can be understood at the level of the student :P
The original question has 4 parts, which are more meaningful in its assessment of concepts and skills. Calculate the probability that
(a) a man chosen at random is taller than 180 cm,
(b) a man and a woman chosen at random are both taller than 180 cm,
(c) of a man and a woman chosen at random neither is taller than 180 cm,
(d) a man chosen at random is taller than a woman chosen at random.
Sometimes, you can refuse to look at questions from school tutorials when they are not well-set and clear in their assessment outcomes.
Thanks.
Cheers,
Wen Shih -
Hi Mr Wee,
Indeed, I agree with you.
Badly set questions should be ignored.
A properly set question will be phrased in a very clear and precise language as those questions in the actual H2 Maths examination papers.
Like the earlier question on probability question on at least one boy, the setter tries purposely to trick students who attempt the question unnecessarily.
A more useful example in statistics could be a question to see whether the students know when to use the correct concept eg to use n x sigma^2 or n^2 x sigma^2 instead of trying to trick the students with the play of words.
Regards.
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Originally posted by Lee012lee:
Question 1
Mr. Lim has two children. At least one of them is a boy. What is the probability that the other one is also a boy?
Steps
(1) Given the condition at least one boy (BB, BG, GB) ie (1/2 x 1/2) x 3
(2) So, given the condition that at least one boy, the probability that the other one
is also a boy ie
P(BB)
= -----------------------
P(BB, BG, GB)
1/2 x 1/2
= ----------------------
( 1/2 x 1/2) x 3
= 1/3
hello, ive always been intriguied by this question ever since i saw a similar question on the wiki page of marylin vos savant. lol. i want to ask, in the above working, why would order of boy or girl matter?
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Originally posted by wee_ws:
Dear Mad Hat,
We can always start a movement to help students understand H2 maths on this forum :)
Sometimes, simple things are made complex because of a fixated mindset that concepts must be taught in a certain way, rather than by the use of skilful means that can be understood at the level of the student :P
The original question has 4 parts, which are more meaningful in its assessment of concepts and skills. Calculate the probability that
(a) a man chosen at random is taller than 180 cm,
(b) a man and a woman chosen at random are both taller than 180 cm,
(c) of a man and a woman chosen at random neither is taller than 180 cm,
(d) a man chosen at random is taller than a woman chosen at random.
Sometimes, you can refuse to look at questions from school tutorials when they are not well-set and clear in their assessment outcomes.
Thanks.
Cheers,
Wen ShihOh, I agree with this. I was just curious what the teacher was thinking such that he could come up with the answer 0.309. What question did he mean to set? But maybe you're right - it's not worth the time. Thank you for showing the original question. I was curious what modifications the teacher made.
As for the question absol just asked above, I think that's a good question, but I'll let Lee answer it!
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Hi Mad Hat,
The conditional probability question is a very simple conditional probability question.
Steps
(1) Draw the tree diagram
(2) Given condition at least one boy ie BB, BG, GB
(3) Require the other child will also be the boy ie BB
P(BB)
(4) Required Probability = --------------------
P(BB, BG, GB)
ie what is the probability of getting BB out of the given condition of (BB, BG,
GB).
PS : Thes types of conditional probability questions appear frequently in LCCI Business Statistics examination questions.
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LCCI Business Stats. Wah, I'd be interested to see what kinds of stats questions they ask on the exams. Re SBS261P: my student is not from RV, but her friend is, so I have access to RV questions. And yes the question about the two boys is similar to the Monty Hall one, but I think the two boys question is 300 years old, whereas Monty Hall is only last century, so maybe Monty Hall is the reincarnation ! P.S. The Monty Hall problem can also be solved with conditional probability, if anyone wants to try. Need to think a bit how to do it, but it's very satisfying to solve Monty Hall with conditional probability.
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LCCI Business Statistics exam question on conditional probability
A motor insurance company classifies its customers into
4 claims categories : Very Good (10%), Good(30%), Average (20%) and Poor (40%). The claims record shows that the percentage in each classification making at least one claim over the previous year.
Claims Category Percentage making at least one claim
Very Good 5%
Good 10%
Average 30%
Poor 60%
Find the probability that, if a customer is selected at random, the customer who makes at least one claim will come from the very good classification.
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Originally posted by Lee012lee:
LCCI Business Statistics exam question on conditional probability
A motor insurance company classifies its customers into
4 claims categories : Very Good (10%), Good(30%), Average (20%) and Poor (40%). The claims record shows that the percentage in each classification making at least one claim over the previous year.
Claims Category Percentage making at least one claim
Very Good 5%
Good 10%
Average 30%
Poor 60%
Find the probability that, if a customer is selected at random, the customer who makes at least one claim will come from the very good classification.
Hi Lee, thanks for sharing. Quite interesting to see the practical orientation of the question.
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Originally posted by absol:
hello, ive always been intriguied by this question ever since i saw a similar question on the wiki page of marylin vos savant. lol. i want to ask, in the above working, why would order of boy or girl matter?
The order doesn't really matter so long as the probability of having a boy and a girl is calculated correctly (as part of the overall calculation). Some find it easier to calculate this by considering the separate cases BG and GB. Each is ¼, so the total is ½.
But if you don't want to bother about the order, you can also calculate:
1 - P(both boys) - P(both girls)
= 1 - ¼ - ¼
= ½.In the end, the prob of having a boy and a girl (in whichever order) is ½. The "ordering" is just to simplify the calculation.
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Originally posted by wee_ws:
Dear Mad Hat,
We can always start a movement to help students understand H2 maths on this forum :)
Ah yes of course, but no no, not on this forum. After all that work then sgforums get paid for it! (ad revenue). Better we get paid if we do all that work
. I sometimes think about setting up a website for H2 maths. I'm quite okay with web programming, can do a proper site with proper functionality and graphics etc.I visited your website and there's a lot of great stuff on it. Did you ever consider turning it into a full-blown website with its own domain name and all the works? One of my ideas is to use videos to show how the teacher thinks while working through a question. (Think out loud while solving the question.) I think this would be helpful to students - a model of what sorts of things they should be thinking about while attempting a question. All that stuff about reasonableness-checking will fit right into the video also. One of the best (and easiest) ways to teach is by modelling. A bit like chess, you know. I play chess and learnt the most by learning what a master player thinks when faced with a position on the board. (I learnt that I was thinking about all the wrong things!)
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Hi Mad Hat,
Thanks for your encouragement!
I created the site to help students learn better. Also, students can reach me even though I do not teach them in person :) Some form of modelling through thinking aloud is also introduced in the articles that I have written.
Yes, videos would be great. I'm working on that :)
Thanks!
Cheers,
Wen Shih -
the reason why i asked why the order of BG or GB matters for that question is because, from a "normal" perspective, if one is asked the same question, that if Mr Lee has at least a boy, what is the probability of the other kid also being a boy, it is more usual to think either mr lee can have 2 boys, or mr lee can have 1 boy and 1 girl. therefore probability is 1/2.
now in terms of the working, if i tell u that i have at least one boy and asks for the probability that my other kid is also a boy, following the working, you will be thinking that possible scenarios are that i have 2 boys, a girl followed by a boy, and finally a boy followed by a girl in which the ordering is dictating whether the girl is an elder/younger sister of the boy. to me its kinda confusing and dosent make sense in real life to think this way.
but after some thoughts if you change it to a coin toss question of head and tails it becomes much easier to absorb :o
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This question is from TYS complex numbers, p. 65, q. 24, in my Dyna edition 2009, yellow cover.
Write down, in any form, all the roots of the equation z5 − 1 = 0.
Hence find all the roots of the equation
(w − 1)4 + (w − 1)3 + (w − 1)2 + w = 0
and deduce that none of them is real.
Find the arguments of the two roots which have the smaller modulus.
This question is quite interesting. I have a question about the last part - find the arguments. The TYS answer is 2π/5 and 3π/5, but my answer is 2π/5 and −2π/5. Mr. Wee (or anyone else), if you can check this for me, I'd be grateful.
Just for comparison, my answer to the first part is z = e^(i 2kπ / 5)
where k = 0, 1, 2, 3, 4.And for second part, I get w = 1 + e^(i 2kπ / 5)
where k = 1, 2, 3, 4.From the Argand diagram, the two roots with the smaller modulus should fall in the first and fourth quadrant. Or did I slip up somewhere?
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Hi Mad Hat,
Yes, your answers of 2π/5 and −2π/5 are correct. Thanks!
It is an interesting question because the concepts of GP (to find the roots of w), transformation (to draw the argand diagram representing the roots of w, based on the roots of z) and geometry (to find the arguments of the two roots) are asked along with those related to complex numbers :)
Thanks again!
Cheers,
Wen Shih -
Gotcha, thank you ! Yes, I think this question will be on the harder end for most students. Anyone wanting to practice flexibility should study this question; see how all the mathematical ideas come together in such a nice way.
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This is from ACJC 2009 prelims, paper 2, Q6, part (ii):
Vehicles arrive randomly at a junction at an average rate of 10 per minute. Traffic counts at this location indicate that 70 percent of these vehicles are cars. Assume that the arrivals of the vehicle follow a Poisson distribution.
(ii) Find the variance of the number of cars that arrive in a 3-hour period.
Answer given:
Let X = number of vehicles arriving at the junction in 3 hours
⇒ X ~ Po(1800)
⇒ Var(X) = 1800
⇒ Var(0.7X) = 0.7 x 0.7 x 1800 = 882.My answer:
Let X = number of cars arriving at the junction in 3 hours
⇒ X ~ Po(1260)
⇒ Var(X) is 1260.Which approach is correct?
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Hi,
You took the proportion of the average rate, so it is scaled to 7 per minute.
However, the average rate is only proportional to the duration according to the conditions when using the Poisson model:
http://www.childrens-mercy.org/stats/definitions/poisson.htm
Thanks for sharing this question for discussion :)
Cheers,
Wen Shih -
Hi Mad Hat,
The question indicates that the number of vehicles arriving follow a poisson distribution. It didn't mention that the number of cars arriving follow a poisson distribution.
Basically, if we take X as the number of cars arriving, and Y as the number of vehicles other than cars arriving, the question tells us that in a 3 hour period,
X+Y~Po(1800)
We can't derive from there that X~Po(1260)
Mr Wee, Mad Hat used 0.7 * 1800 = 1260 cars arriving.